A point on a rotating wheel has an angular position given by ? = 6.0 + 4.0 t 2 +

Question

A point on a rotating wheel has an angular position given by ? = 6.0 + 4.0t2 + 4.0t3, where ? is in radians and t is in seconds. The point is located a distance 43 cm from the axle of the wheel.

Answers must have units!
(a) At t = 0, what is the point's angular position?


(b) At t = 0, what is the point's angular velocity?


(c) What is the point's angular velocity at t = 4.0 s?


(d) What is the tangential speed of the point at t = 4.0 s?


(e) What is the radial speed of the point at t = 4.0 s?


(f) Calculate the point's angular acceleration at t = 2.0 s.


(g) What is the magnitude of the tangential acceleration of the point at t = 2.0 s?


(h) What is the magnitude of the radial (centripetal) acceleration of the point at t = 2.0 s?

Note that the radial acceleration is much larger than the tangential acceleration. As the angular speed increases, the forces required to keep the wheel intact increase dramatically, and at high enough speeds the wheel will explode.
(i) Is the angular acceleration constant?

yesno     not enough information

Explanation / Answer

here,

theta = 6 + 4t^2 + 4 t^3

differentiating the equation for w

w = 8 * t + 12 * t^2

a)

at t = 0 s

theta = 6 + 4*)^2 + 4 * 0^3

theta = 6 rad

b)

at t = 0 s

w(0) = 8 * t + 12 * t^2 = 0 rad/s

c)

at t = 4 s

w(4) = 8 * 4 + 12 * 4^2 = 224 rad/s

d)

the tangential speed of the point at t = 4.0 s , v = r * w(4)

v = 96.3 m/s

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